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Mplus Discussion > Confirmatory Factor Analysis >
 Anonymous posted on Friday, April 23, 2004 - 5:45 am
I have recently run a confirmatory factor analysis with several constructs related to the development of antisocial behavior. Although fit indicies and chi-square change statistics indicate that the four factor solution proposed is the best fitting model, the factors are highly correlated (.75-.84). Will statisticians question the independence of these constructs because of their high correlation? In addition, do you run into the same problems in SEM with multicollinearity when including highly correlated factors in a regression together?
 bmuthen posted on Friday, April 23, 2004 - 6:51 am
I don't think this is problematic. Factors are often naturally correlated, not independent. Multicollinearity is always a risk. One way around this is to postulate a second-order factor behind highly correlated factors.
 Anonymous posted on Wednesday, April 28, 2004 - 5:22 am
Thanks for the info. A few follow-up questions. Do you have reccomendations on how to check for multicollinearity in Mplus 3.0? In addition, when a second-order factor is used to handle the issue, do you simultaneously regress the DV onto the second order factor and all first order factors, or do you simply regress on the DV onto all first order factors without including the second order factor?
 bmuthen posted on Thursday, April 29, 2004 - 6:15 pm
I think multicollinearity can be checked just like in regular regression with observed variables. With a 2nd-order factor, you can simply regress the DV on that factor only.
 Anonymous posted on Sunday, November 21, 2004 - 6:18 pm

I just have a question about the correlation between factors in a confirmatory factor analysis.
I wonder if a rotation is used when doing a CFA. If yes, is it promax?
If no, how are calculated the correlations between factors?
Another question: if the observed variables are categorical, does it make a difference about the correlation between factors?

 Linda K. Muthen posted on Monday, November 22, 2004 - 8:34 am
CFA does not use a rotation. The covariances are estimated as part of the model. Whether the factor indicators are contiuous or categorical, the factor is continuous.
 Boliang Guo posted on Saturday, September 03, 2005 - 11:32 am
Hi Linda,
how to set the factors correlation is 1? thank for your your kind attention.
 Michael J. Zyphur posted on Monday, September 05, 2005 - 9:08 am
I know the question was directed at Linda, but the answer is:

f1 with f2@1.0

where f1 = factor 1 and f2 = factor 2
 Linda K. Muthen posted on Monday, September 05, 2005 - 2:18 pm
Note that this is only a correlation if the metric of the factor is set by fixing the factor variances to one. If the metric of the factors is fixed by setting one factor loading to one, then this is a covariance.
 yufang posted on Thursday, October 06, 2005 - 11:39 am
Does anyone know references that recommends how large a correlation between factors is considered moderate or high? and at which point should we check for multicollinearity?
 bmuthen posted on Saturday, October 08, 2005 - 11:55 am
I do not. - Others?
 yang posted on Monday, April 17, 2006 - 12:30 pm
Why the covariance matrix and correlation matrix are identical for factors in CFA? Thanks.
 Linda K. Muthen posted on Monday, April 17, 2006 - 1:43 pm
Your factor variances must be one.
 yang posted on Friday, June 23, 2006 - 6:53 am
I ran a CFA and got some correlation coefficients among the factors with absolute values greater than 1. Thank Linda for telling me that this means the corresponding factors are not statistically distinguishable. Now I have another question: how can the correlations coefficients among the factors have absolute values greater than 1? Are we using some different formula to calculate these coefficients? Shouldn¡¯t the correlation coefficients should range from -1 to 1? Thanks a lot.
 Linda K. Muthen posted on Friday, June 23, 2006 - 8:35 am
Correlations are not calculated using a formula. The correlations are estimated as part of the model. When variables correlate one, model estimation is thrown off and values greater than one can occur. This is why the results are inadmissible in this case.
 yshing posted on Tuesday, November 14, 2006 - 9:53 am
I'm running a multiple-group CFA with 2 factors. For theoretical reason I don't want to fix the variance of the latent construct to one. For one of my groups I would like to fix the correlation of the two latent constructs to 1. I understand that a nonlinear constraint is needed in the model. How can I do this?

Thank you for your help.
 Linda K. Muthen posted on Tuesday, November 14, 2006 - 1:32 pm
See MODEL CONSTRAINT in the user's guide. You can create a new parameter that is the covariance diviied by the product of the two standard deviations and make the new parameter equal to one.
 Sigbert Klinke posted on Friday, December 01, 2006 - 7:27 am

i'am trying to make a confirmatory factor analysis from categorical variables only:

f1 BY fte1*0.600 fte2*0.526 fte6*0.745 fte7*0.654 fte8*0.690 fte10*0.639
fte11*0.655 ftt16*0.409 ftt17*0.460 ftt18*0.463 ftf18*0.408;
f2 BY ftt19*0.405 ftf1*0.745 ftf2*0.749 ftf3*0.628 ftf8*0.472 ftf9*0.428
ftf13*0.484 ftf17*0.402;

f1@1 f2@1;
f1 WITH f2@0;

Under this model I would assume that the factors have variance 1 and uncorrelated. However, the saved factor scores are still correlated and have a variance unequal to one (around 0.5).
Do I have to specify more?

Thanks for any help

 Linda K. Muthen posted on Friday, December 01, 2006 - 9:45 am
Factor scores are not identical to the factors in the estimated model. Deviations can occur when the factors do not have high factor determinacties. This is one reason that it is better to work with factors in a simultaneous model rather than work with factor scores.
 Sigbert Klinke posted on Monday, December 04, 2006 - 5:50 am

so, what do I save with the scores? The deviations are pretty large for a two factor model. Can it be, because the model is bad (CFI 0.171, TLI 0.202, RMSEA 0.243)? I wanted to test a 2-factor model, before I go to the full 9-factor model.
Can I check somewhere how many iterations the CFA needed? Maybe the model has not converged, it took some minutes to finish; although estimating each factor separately was very quick.
Since we want to estimate a SEM or a Multilevel model with these factors with another program I need reliable factor scores, if possible.

Thanks in advance

Sigbert Klinke
 Linda K. Muthen posted on Monday, December 04, 2006 - 8:25 am
If the model did not converge, it would say so in the output. I think the differences can be attributed to a poorly fitting model.
 Daniel Shen posted on Tuesday, March 13, 2007 - 8:19 am
Prof. Muthen,

In CFA, can we obtain standard errors of factor correlation estimates, in order to build a 95% CI? Thanks,

 Linda K. Muthen posted on Thursday, March 15, 2007 - 4:36 pm
You can do this in two ways. You can set the metric of the factors by setting the factor variances to one. Then you obtain correlations among the factors rather than covariances. Or you can use MODEL CONSTRAINT to create a correlation from the covariance.
 Ioanna Vrouva posted on Sunday, November 16, 2008 - 7:48 am
I would be very grateful for your advice:
1) Assuming I do a CFA and want to test the orthogonal model, would
"f1 WITH f2@0" be the correct syntax to use?
2) Is it possible to obtain CFI, TLI, RMSEA, WRMR values for the the independence model (standard control in CFA) and what would the relevant syntax be? (I have 23 indicators that are ordinal/categorical data and 2 latent variables)
I tried
f1 BY v1
f2 BY v2
f23 BY v23
but it obviously didn't work

many thanks
 Bengt O. Muthen posted on Sunday, November 16, 2008 - 10:15 am
1) Yes.

2) The independence model is called the "baseline model" in Mplus and is automatically included so that CFI etc can be reported - see output.
 Ioanna Vrouva posted on Wednesday, November 19, 2008 - 11:27 am
Thanks very much. Unfortunately, I cannot see any CFI/TLI/RMSEA/WRMR values for the baseline model in the output. I can see the relevant chi-square, df and p value though. Where should I look for the CFI/TLI/RMSEA/WRMR of the baseline model?
many thanks
 Bengt O. Muthen posted on Wednesday, November 19, 2008 - 6:29 pm
You don't get CFI etc for the baseline model because that would just compare the baseline model to itself. The baseline model is merely used to be able to compute CFI.

If you are interested in chi-2 fit for the baseline model, you would have to specify it yourself in the MODEL command:

y1-yp with y1-yp@0;

where p is the number of variables. This will then get you a chi-2 test saying how well (probably how very badly) the baseline model fits relative to an unrestricted model. Again, CFI is not relevant since both the H1 and H0 models are the baseline model.
 Ioanna Vrouva posted on Sunday, November 30, 2008 - 5:07 am
This is very clear, thanks very much.
 Kihan Kim posted on Wednesday, January 28, 2009 - 10:11 pm
Dear Dr. Muthen,

I'm testing a two-factor CFA model with 7 items (4 items for F1, and 3 items for F2).

I wanted to perform a chi-square difference test between (M1) two-factor CFA (factors are allowed to correlate), and (M2) two-factor CFA with inter-factor correlation fixed at 1.

When I run M2, I'm keep receiving the following convergence problem. I looked over the User Manual regarding "Convergence Problems," and am still not sure what I should try. Could you help me resolving this problem?


 Linda K. Muthen posted on Thursday, January 29, 2009 - 8:30 am
Unless you have set the metric of the factor by freeing all factor loadings and fixing the factor variance to one, you are fixing the covariance to one not the correlation.

Fixing a parameter to an incorrect value may cause convergence problems. If you want to test if a covariance or correlation is equal to one, use MODEL TEST. See the user's guide for further information.
 Kihan Kim posted on Friday, January 30, 2009 - 9:04 am
Thank you for your answer. I was able to fix the inter-factor correlation to 1 by freeing all factor loadings and fixing the factor variance to one.

I was also trying to use MODEL TEST, but I'm still not clear how to use it. Could you suggest a command so that I can set the inter-factor correlation to 1 using MODEL TEST command for the following MODEL command of 2-factor CFA?

Model: f1 by id1 id2 id3 id4;
f2 by fit1 fit2 fit3;
 Linda K. Muthen posted on Friday, January 30, 2009 - 9:46 am
The following code tests if the factor correlation is one. It does not fix it to one.

MODEL: f1 BY id1* id2 id3 id4;
f2 BY fit1* fit2 fit3;
f1 WITH f2 (p1);

0 = 1 - p1;
 tina freyburg posted on Tuesday, February 24, 2009 - 7:31 am
Dear Linda & Bengt,

I wonder how you can get the significance level of oblimin factor correlations. Would be great if you helped me.

Thanks a lot in advance,

 Linda K. Muthen posted on Tuesday, February 24, 2009 - 9:24 am
You can't get these with TYPE=EFA but you can get them with the new EFA using the MODEL command. See the Version 5.1 Examples and Language Addendums on the website with the Mplus User's Guide.
 Tracy Witte posted on Thursday, March 04, 2010 - 9:57 am
I am running a CFA with the WLSMV estimator. Similar to the original person on this thread, the correlation between my factors is very high (in my case, approximately .92). However, when I use the difftest procedure to compare the two factor to the one-factor model, the results show that the fit of the model is significantly worsened when I specify a one-factor model. Is my next step to determine if the factors have differential predictive validity? It seems unlikely with such a high correlation between the two of them. From my reading of the above thread, it looks like I should specify a higher-order factor and regress the DV onto lower-order factors. Is this correct? (thank you!)
 Linda K. Muthen posted on Thursday, March 04, 2010 - 10:13 am
I would try an EFA to see if your CFA model is even in the ballpark for these data.
 Matthew Diemer posted on Sunday, March 28, 2010 - 6:35 pm
I want to confirm I am interpreting my output correctly, as I am unsure exactly how the unstandardized vs. STDYX output is calculated within M+.

Within a larger model, I have multicollinearity between two latent predictors (Parental Support and Peer Support) of the latent outcome variable - educational expectations.

The issue I have is that the standard errors (SEs) associated with these direct effects are much much larger in the unstandardized output than in the standardized STDYX output.

I know inflated SEs are a sign or symptom of multicollinearity and have addressed this by constraining the paths of peer and parental support --> educational expectations as equal. (I'll spare the details).

Even after doing so, why are the SEs associated with these two direct effects so much larger in the unstandardized section of the output than in the STYX output?
 Linda K. Muthen posted on Monday, March 29, 2010 - 8:26 am
See the STANDARDIZED option in the user's guide and Technical Appendix 3 on the website for information about standardizations used in Mplus.

It sounds like your efforts to get around your multicollinearity are not succeeding. I suggest using only one of the variables involved.
 Antti Kärnä posted on Monday, April 05, 2010 - 9:37 am
Hi, I would like to test whether factor correlations are equal to each other in CFA, and I have defined in Mplus:
F1 with F2 (1);
F2 with F3 (1);
F3 with F4 (1);
F5 with F6 (1);
I got the output where all covariances are 0.044, but correlations from STDYX standardization range from 0.338 to 0.512.

LR test from EQS: Chi^2(3)=6.89, P=.07
LR test from Mplus: Chi^2(3)7.80, P=.05

Have I performed the right test in Mplus, if I want to replicate the results from EQS? Thanks in advance.
 Linda K. Muthen posted on Tuesday, April 06, 2010 - 8:03 am
The difference in results between Mplus and EQS is likely due to Mplus using n and EQS using n-1 for the chi-square computation. This difference shows up for small samples which I assume you have.
 Delforterie posted on Friday, March 18, 2011 - 6:30 am

In this post I read that when the correlation between factors is greater than 1, this means the corresponding factors are not statistically distinguishable. But does it also mean that the correlation is actually 1.00, and the model is simply estimating a somewhat higher correlation?

Additionally, does this mean that when I have 2 factors with a correlation of 1.16, I can assume a one-factor model fits the data better?
When I did a Chi-square difference test, it indicated that the two factor model was better.
 Bengt O. Muthen posted on Friday, March 18, 2011 - 8:44 am
A correlation estimate of 1 means that the factors are indistinguishable. A correlation estimate higher than 1 means that the model does not make sense for the data because correlations should not be higher than 1. So even if chi-square says that two factors fit better, you should not choose that model. Instead, another model should be explored.
 Vandita Vasudevan posted on Friday, May 27, 2011 - 1:27 pm
I am new to MPlus and I am still learning the codes to run the analysis. I am running a five factor CFA model and wanted to remove non significant correlations between 2 latent factors. Could you please tell me how this is done in MPLUS.
Thanks for your help.
 Linda K. Muthen posted on Friday, May 27, 2011 - 2:30 pm
If you want to fix the factor covariance to zero, you do so by

f1 WITH f2@0;
 Vandita Vasudevan posted on Friday, May 27, 2011 - 2:44 pm
model: fc by f1 f6 f11 f16;
ex by f7 f17;
cf by f3 f8 f13 f18;
au by f4 f9 f14 f19;
lf by f5 f10 f15 f20;
lf with fc@0;

Does this look correct?
 Linda K. Muthen posted on Friday, May 27, 2011 - 2:58 pm
 Jan Breitsohl posted on Monday, October 03, 2011 - 4:26 am

I have a simple question:

I am doing a 7 factor CFA and my model results indicate some insignificant factor covariances (i.e. with statements of 0.443 and 0.177).
What do you recommend to do about it?

What is the effect if I set them to 0 and what is the effect if I simply do nothing about it?

Global fit indices are all very good and all BY-statements are statistically significant.

Thanks in advance,

 Linda K. Muthen posted on Monday, October 03, 2011 - 5:31 am
I would not fix them to zero. I would leave them as is.
 Xiaolu Zhou posted on Tuesday, November 08, 2011 - 1:22 pm
I am a new user. Could you help me with my CFA syntax? My data is binary data. I am not sure if my syntax is correct to check the correlation between the 2 factors with this syntax. My syntax is:
TITLE: c scale

NAMES ARE country c16 c17 c18 c19 c20 c21
c22 c23 c24 c25 c26 c27 c28 c29

USEVARIABLES ARE c16 c18 c19 c20 c21
c22 c23 c26 c27 c30;

CATEGORICAL ARE c16 c18 c19 c20 c21
c22 c23 c26 c27 c30;

MISSING are all (999);

MODEL: b BY c16*
c18 c23 c26 c27 c30;

v BY c19* c20 c21 c22;



b With v;


OUTPUT: standardized MODINDICES (3.5);

For the output part, I have two questions: 1.where I can find the correlation between the two factors? 2. for the model fit, if chi-square, CFI and RMSEA is good, only TLI is less than .95, can we still call the model fit is acceptable? Many thanks!
 Linda K. Muthen posted on Tuesday, November 08, 2011 - 1:54 pm
The correlation is in the results under b WITH v. If most fit statistics show good fit, that should be acceptable.
 Xiaolu Zhou posted on Wednesday, February 01, 2012 - 10:19 am
Hi Linda,

I have another question about the CFA of binary data: I found some non-significant thresholds. What does these mean? Do they matter to my model? If they matter, what should I do with them? Thanks a lot!
 Linda K. Muthen posted on Wednesday, February 01, 2012 - 11:20 am
Thrsholds are used in the computation of probabilities and to test for measurement invariance. I would not be concerned with their significance.
 Sarah  posted on Friday, October 05, 2012 - 12:49 pm

I have a quick and probably very easy question.

In my SEM model I wish to obtain the correlations between my latent factors. I used the "tech4" option which provides such correlations. But how do I know if the correlations are significant? Put differently how can I obtain the level of significance of the correlations?

Thank you very much.

 Linda K. Muthen posted on Friday, October 05, 2012 - 12:52 pm
We don't provide this. You could define the correlations in MODEL CONSTRAINT and get a standard error that way.
 sailor cai posted on Wednesday, November 07, 2012 - 2:35 am
Hi Linda,

One question about standardization:

I am testing latent interaction in a SEM model. The direction between two factors is two-way. So,how can I standardize correlation coefficients?


F4 on F1 F2 F3;
F1 with F2 F3;
F2 with F3;

So, how to standardize, say, r12?


 Linda K. Muthen posted on Wednesday, November 07, 2012 - 9:48 am
If f1-f4 are factors defined using the BY option, You can ask for STD in the OUTPUT command. I don't see any interaction in the MODEL command.
 sailor cai posted on Wednesday, November 07, 2012 - 7:40 pm
Hi Linda, thanks for the direction. Sorry for not giving the whole syntax. My original commants are:


F1 by b1 b2 b3 b4; F2 by l1 l2; F3 by s1 s2 s3 s4 s5 s6;F4 by t1 t2 t3 t4;

F4 on F1 F2 F3 F1xF2; F1 with F2 F3; F2 with F3; F1xF2 | F1 XWITH F2;

As the TYPE=random is used, therefore the STD cannot be asked for. Is there an equation to calculate the standardized correlations?

Thanks again!
 Linda K. Muthen posted on Thursday, November 08, 2012 - 9:36 am
This is not possible if numerical integration is required.
 sailor cai posted on Thursday, November 08, 2012 - 9:34 pm
By "not possible", do you mean asking for standardized correlation is only not possible:

1) using Mplus commands?
2) yet to be developed in the literature?
3) theoretically not possible?

Your further clarification will be appreciated!
 Linda K. Muthen posted on Friday, November 09, 2012 - 11:16 am
1. Yes.
2-3. See the FAQ Latent Variable Interactions on the website
 shumail paracha posted on Friday, May 31, 2013 - 12:42 am
Dear Sir;

please guide me....
i have utilized second order factor model..where

F1 BY y1 y2 y3;

F2 BY y4 y5 y6;

F3 BY y7 y8 y9;

F4 BY F1 F2 F3;

F5 BY x1 x3 x4 x5;

F6 BY v1 v2 v3 v4;

F5 ON F4;

F6 ON F5;


Fit indices are CFI=.912, TLI=.90, RMSEA=.038, SRMR=.05

But problem is that F4 which is comprised of three factors, shows non-significant relationship with F2,

I have checked that scale correlation with each subscale and get highly significant correlation…yes no doubt structural equation modeling better deals with measurement errors and consider residuals covariance as well….but please tell me is there any way to come out of this problem….

F1 has .832 standardized estimates with F4

F2 has -.022 non sig correlation with F4…..its residual variance estimate is 1 n significant….

F3 has .899 standardized estimates

now how i would report one factor is non significant....but model fit indices are good enough
 Bengt O. Muthen posted on Friday, May 31, 2013 - 8:15 am
This general SEM question is better suited for SEMNET.
 Kwan K. posted on Sunday, June 30, 2013 - 11:27 am
Hi Linda,

I saw that you previously replied to Sarah (question above) that Mplus does not provide the level of significance of the correlations.

However, I have found several articles using Mplus and reporting the level of significance of the factor correlations. I was wondering if there is any other way to get it or I missed something here.


 Linda K. Muthen posted on Sunday, June 30, 2013 - 7:15 pm
These are now available in TECH4 of the OUTPUT command in most cases.
 Kwan K. posted on Monday, July 01, 2013 - 1:21 am
Thank you Linda.
Do you mean these are available in the latest version of Mplus? since it apparently does not work for my Mplus VERSION 6.11.


 Linda K. Muthen posted on Monday, July 01, 2013 - 5:20 am
Yes, they are available in the latest version of Mplus.
 Hugo Cogo-Moreira posted on Tuesday, December 17, 2013 - 2:00 pm
Dear Professors,
I am trying to check if there is Multicollinearity with my high two correlated factors (.86) in a regression. As suggested, I simply regressed the both factors on a second-order factor. I obtained
VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE, where my second-order factor and a DV had a correlation greater than 1.
Is it a indicator of multicolinearity?
All best,
 Bengt O. Muthen posted on Tuesday, December 17, 2013 - 3:39 pm
You need more than two factors to define a second-order factor is a solid way. It isn't identified without restrictive specifications.

A correlation of 0.86 most likely gives rise to multicollinearity problems.
 Eric Thibodeau posted on Monday, February 23, 2015 - 3:47 pm

I want to make sure what I'm doing is the correct way of correlating factors to each other and factors to observed variables.

I ran an SEM and for publication I want to report a table of correlations between all variables used (observed and latent).

For factor correlations the syntax is:

f1 with f2:

For factor and observed correlations:

f1 with age;

Then I examine the respective STDYX "with" estimates ?

 Bengt O. Muthen posted on Tuesday, February 24, 2015 - 7:50 am
Yes, but make sure you don't change your SEM model by adding these correlations. You find latent variable correlations in TECH4. If you want correlations between latent and observed variables not included in TECH4 you have to put a factor behind the observed variables.

Typically, however, you report the model parameters, not these correlations.
 Emily Sherman posted on Sunday, April 26, 2015 - 11:39 am
I have a simple question about including multiple correlation (WITH) statements in a model. I have a latent factor that I am then correlating with the five domains of personality modeled as separate indicators. I have run five separate models with one personality domain correlated with the latent factor at a time, but I want to correlate all 5 of them concurrently with the latent factor. This runs fine and the values are similar to those acquired when I run the separate models. My question is regarding the concurrent model- are these correlations independent of each other or will they be controlling for the other indicators that are being correlated with the latent factor at the same time?
 Bengt O. Muthen posted on Sunday, April 26, 2015 - 4:19 pm
They will be regular correlations if you also allow correlations among your 5 personality domain variables so that this part of the model is just-identified (saturated).
 Emily Sherman posted on Sunday, April 26, 2015 - 5:31 pm
Thanks- otherwise would they be semi-partial correlations..or? (If I don't allow them to be correlated that is)
 Bengt O. Muthen posted on Monday, April 27, 2015 - 11:09 am
No, in that case they are just ignorable due to a misfitting model.
 M.O. posted on Friday, May 01, 2015 - 11:28 pm
I am comparing 1 and 2 factor models for factor mixture analysis. In the 2 factor model, factor correlation turned out to be 1. I suspect this means factor 1 and 2 are statistically not distinguishable, and 2 factor model should not be considered.

However, AIC and BIC are smaller in 2 factor model than 1 factor model (AIC, 10111 vs 10012; BIC, 10273 vs 10189). Does this mean that 2 factor model is better model for my data? Is there anything I could do to improve model?

Just to clarify, here is the input statement for 2 factor model.
Thanks a lot for your advice,

FILE = '35itemN888.dat';
NAMES = u1-u35;
USEVARIABLES = u14-u17 u30-u32;
CATEGORICAL = u14-u17 u30-u32;
f1 by u14-u17;
f2 by u30-u32;
Estimate S.E. Est./S.E. P-Value
F1 1.000 0.000 ********* 0.000
 Bengt O. Muthen posted on Saturday, May 02, 2015 - 10:52 am
You could check which factor loadings are big and for pairs of items with large loadings you can replace the factor with WITH statements for those pairs of items. Use Parameterization=Rescov. See the new article on our website:

Asparouhov, T. & Muthen, B. (2015). Residual associations in latent class and latent transition analysis. Structural Equation Modeling: A Multidisciplinary Journal, 22:2, 169-177, DOI: 10.1080/10705511.2014.935844
 M T posted on Tuesday, January 26, 2016 - 1:12 am
In CFA/full SEM model analysis of a second-order construct, is it correct to include covariances between error terms of the items of two latent factor?

For example, I have a second order construct of internal social capital and (structural, relational and cognitive) as the latent indicator of internal social capital. Then structural is measured by s1, s2,s3,s4, relational is measured by r1,r2,r3,r4 and cognitive is measured by c1, c2, c3,c4. When doing analysis and to enhance the fitness of model, can I consider the covariance between for example s1 and r3?

If I do so, then there will not be any discriminant validity of these latent constructs (structural, relational and connive. Is this correct? Should not discriminant validity be assessed for these latent constructs?
 Bengt O. Muthen posted on Wednesday, January 27, 2016 - 3:10 pm
Q1. Yes if you include only some.

Q2. Yes.

Q3-4. I would think so, but get more opinions on SEMNET.
 M T posted on Thursday, March 17, 2016 - 4:53 am
Is it possible to have the mean, standard deviation, correlation of constructs, Average Variance Extracted and Composite Reliability in MPlus? What is the command please? I am runing CFA and every time I have had to transfer the data to SPSS to calculate these. It would be great if I can perform these analysis in MPlus.
 Linda K. Muthen posted on Thursday, March 17, 2016 - 11:07 am
You can obtain the mean, variance, and correlations using TECH4 of the OUTPUT command. Mplus does not provide the other information.
 Laurie Hawkins posted on Monday, July 25, 2016 - 10:56 am
I am a new user and ran a CFA for 1 factor with 4 items and got a non-significant Chi-Square of 0.288 (p=.5917) with 1 degree of freedom.
My RMSEA is 0.000
and my CFI/TLI are:
CFI 1.000 TLI 1.007

It looks to me as if there is likely something wrong with the model but I am at a loss as to what is wrong or how to fix it. Please advise.
 Linda K. Muthen posted on Monday, July 25, 2016 - 12:03 pm
Your observed variable correlations are likely low making it hard to reject your model.
 Laurie Hawkins posted on Monday, July 25, 2016 - 12:06 pm
Is there anything I can do to fix this? Or does it simply mean that I cannot use that model?
 Linda K. Muthen posted on Monday, July 25, 2016 - 2:50 pm
You just don't have much power to reject the model which affects the fit statistics.
 Edwin Wouters posted on Wednesday, September 07, 2016 - 10:02 am

We are still busy with the stigma factors. two of the stigma scales have a rather high correlation (0.850) and we have seen that other articles then do an analysis to prove that these are two separate constructs:

The perform a chi square diff test between (1) a model with the two scales and (2) a model with the two scales and the two latent constructs correlating perfectly. If the model with the perfect correlating factors is not significantly better, then the two scales are measuring differnt concepts. Am I correct in this?

Now we tried this with two stigma scales and modelled:

Model A:
ExtStigma by ST1 ST2 ST3;
IntStigma by ST3 ST4 ST5;

Model B:
ExtStigma by ST1 ST2 ST3;
IntStigma by ST3 ST4 ST5;
ExtStigma WITH IntStigma@1;


Are we doing something wrong?
Many thanks, Edwin
 Linda K. Muthen posted on Wednesday, September 07, 2016 - 1:41 pm
Model B fixes the covariance between the two factors to one not the correlation. If you want to test the correlation, you should set the metric of the factors in the factor variance, for example,

ExtStigma by ST1* ST2 ST3;
IntStigma by ST3* ST4 ST5;
ExtStigma WITH IntStigma (p1);

It may be better to test the correlation of one using model test.

0 = 1 - p1;
 Sonja Kumlander posted on Monday, October 02, 2017 - 6:29 am

I am trying to run a CFA in order to check if the six-factor model fits out data. How can I explore whether there's (too much) multicollinerity among my latent variables? Thank you in advance!
 Bengt O. Muthen posted on Monday, October 02, 2017 - 5:01 pm
There is no automatic checking for this (in Mplus). You may want to ask on SEMNET.
 Hillary Gorin posted on Tuesday, November 14, 2017 - 2:49 pm

How much can factors correlate in CFAs?
At what value do fit statistics become questionable? I assumed that above 0.90 may be problematic but the following handbook states above 0.85. Do you know if there is a firm cutoff?


Hillary Gorin

[A Handbook on SEM 2nd Edition
Zainudin Awang - Universiti Sultan Zainal Abidin -Ch. 3: VALIDATING THE MEASUREMENT MODEL: CFA]
 Bengt O. Muthen posted on Wednesday, November 15, 2017 - 10:24 am
No cutoff value. You may want to post this on SEMNET.
 Hillary Gorin posted on Thursday, November 16, 2017 - 10:17 am
Thank you for your help!

Why is there no cutoff value?
 Bengt O. Muthen posted on Thursday, November 16, 2017 - 3:39 pm
It depends on many different features I would think and therefore a fixed cutoff doesn't make much sense.
 Alissa Mahler posted on Tuesday, December 05, 2017 - 4:25 pm

I am trying to test whether a 1-factor or 2-factor model best fits a series of indicators. I was told that if I run a model constraining the variance between to two factors to "1" (and compare it to the unconstrained model), this would test whether a 1 or 2 factor model is more appropriate.

I'm wondering if this seems correct? If it is, when I try to run this test I cannot get the model with the constrained covariance to run. As indicated by the threads above, I've freed all factor loadings and fixed the factor variance to 1, but still get the following error message:


Thanks in advance for your assistance.
 Bengt O. Muthen posted on Tuesday, December 05, 2017 - 5:23 pm
See our 2 FAQs

Testing of factor corr=1

Testing 1 versus 2 factors
 jafar bakhshaie posted on Saturday, April 21, 2018 - 3:25 pm
I have an observed and a latent exogenous variable that I have set them to be correlated in my model.

For the class assignment, they require me to also present all of my parameters within the matrices.

I was wondering what type of matrix is suitable to report the correlation between an observed and a latent variable (I am already using these 4 matrices that none of which seem to be suitable for this purpose: 1. Observed variable measurement model: 2. Factor-loading matrix (Lambda: Λ); 3. Latent variable structural relationship: Latent-regressions (Beta, Β); 4. Observed variable residual covariance matrix (Theta, Θ); Latent variable (residual) covariance matrix (Psi, Ψ) ).

Thanks in advance!
 Bengt O. Muthen posted on Sunday, April 22, 2018 - 10:19 am
Check Tech1 to see where this ends up. by default in some cases, Mplus puts a factor behind an observed variable which then determines the matrix that is used.
 Sara Aguilar-Barrientos posted on Monday, June 25, 2018 - 3:06 pm
I obtained a very high correlation between two latent constructs. It is a multi-group analysis. In model one, variances were constrained at 1, and they were freed in model 2. In model 1, I obtained a correlation of.922. How to deal with this situation?
 Bengt O. Muthen posted on Monday, June 25, 2018 - 3:39 pm
I gave an answer for this in my other reply to you.
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